Evaluate $\int x\cos (6x) \mathrm dx$? $$\int x \cos(6x)\, \mathrm dx$$
I have many similar problems to do, but I keep getting stumped on what to do with what resides inside the parenthesis as opposed to an exponent or something in front of the problem say either $\cos^6 (x)$ or $6 \cos (x)$. What should I be doing differently to solve this integral which has the $6x$ evaluated within cosine?
 A: integrating by parts : we know $$ d(uv) = u \ dv + v \ du $$ 
$$ \Rightarrow \int d(uv) = \int u \ dv + \int v \ du \Rightarrow uv =  \int u \ dv + \int v \ du \Rightarrow \int u \ dv = uv - \int v \ du  $$
we have the integral $$ \int x\cos(6x) \ dx $$
by letting $$ u = x \quad , dv = \cos(6x) \ dx $$
$$ \Rightarrow du = dx \quad , v = \frac{\sin(6x)}{6} $$
so we integrate by parts and we'll get : 
$$ \int x\cos(6x) \ dx = uv - \int v \ du = \frac{x\sin(6x)}{6} - \frac{1}{6}\int \sin(6x) \ dx = \frac{x\sin(6x)}{6} + \frac{\cos(6x)}{36}  + C $$ 
A: Whenever you have the product of:


*

*Something you know how to differentiate (e.g. $x$), and...

*Something you know how to integrate (e.g. $\cos(6x)$)


...you should use integration by parts.

For evaluating $\int \cos(6x)\,dx$, we use a $u$-substitution; let $u=6x$.  This means $du = 6\,dx$.  Now we have:
$$\int \frac{\cos(u)}{6}du$$
A: First do a u-sub on the 6x. The integral then becomes of the form (u)(cosu) with a factor upfront. Then apply Integration By Parts.
